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Relativistic mechanics--Scalars-- 4-vectors -- 4-D velocity-- 4-momentum, rest mass -- conservation laws-- Collisions-- Photons and Compton scattering-- Velocity addition (revisited) and the Doppler shift-- 4-force
1. Scalars
A scalar is a quantity that is the same in all reference frames, or for all observers.
It is an invariant number.
E.g., )lengthproper ,(),timeproper ,(,)( 2restrest llts
But the time interval ∆t, or the distance ∆x between two events, or the length l separating two worldlines are not scalars: they do not have frame-independent values.
2. 4-vectors ),,,( zyxtcx
This 4-vector defined above is actually a frame-independent object, although the components of it are not frame-independent, because they transform by the Lorentz transformation.
E.g., in 3-space, the Different observers set up different coordinate systems and assign different coordinates to two points C and L, say Canterbury and London.
--They may assign different coordinates to the point of the two cities
--They agree on the 3-displacement r separating C and L., the distance between the two points, etc.
With each 4-displacement we can associate a scalar the interval (s)2 along the vector. The interval associated with the above defined 4-vector is
2222222 )()()()()()()( zyxtcrtcs
Because of the similarity of this expression to that of the dot product between 3-vectors in three dimensions, we also denote this interval by a dot product and also by 22222
)()()()( zyxtcxxx
and we will sometimes refer to this as the magnitude or length of the 4-vector.
--We can generalize this dot product to a dot product between any two 4-vectors
zzyyxxtt
zyxtzyxt
bababababa
bbbbbandaaaaa
:),,,(),,,(
--When frames are changed, 4-displacement transform according to the Lorentz transformation, and obeys associativity over addition and commutativity :
abbacabacbai
;)()
ii) A 4-vector multiplied or divided by a scalar is another 4-vector
3. 4-velocity
In 3-dimensional space, 3-velocity is defined by dt
rrlimˆ
0
d
tv
t
where ∆t is the time it takes the object in question to go the 3-displacement ∆ r.
However, this in itself won't do, because we are dividing a 4-vector by anon-scalar (time intervals are not scalars); the quotient will not transform according to the Lorentz transformation.
??dt
xdu
Can we put the 4-displacement in place of the 3-displacement r so that we have
The fix is to replace ∆t by the proper time ∆ corresponding to the interval of the 4-displacement; the 4-velocity is then
twherex
u
:lim0
),,,(),,,(),,,( zyx vvvcdt
dz
dt
dy
dt
dx
dt
dtc
d
dz
d
dy
d
dx
d
dtc
d
xdu
),( , zyx vvvwhere are the components of the 3-velocity dt
rˆd
v
x
Although it is unpleasant to do so, we often write 4-vectors as two-component objects with the rest component a single number and the second a 3-vector. Inthis notation
)ˆ,( vcu
--What is the magnitude of u
The magnitude must be the same in all frames because2u
is a 4-vector.u
Let us change into the frame in which the object in question is at rest.
In this frame 1and)0,0,0(ˆfor)0,0,0,( vcu
cuorcu 22
It is a scalar so it must have this value in all frames. You can also show this by calculating the dot product of
You may find this a little strange. Some particles move quickly, some slowly, but for all particles, the magnitude of the 4-velocity is c. But this is not strange,because we need the magnitude to be a scalar, the same in all frames. If you change frames, some of the particles that were moving quickly before now move slowly, and some of them are stopped altogether. Speeds (magnitudes of3-velocities) are relative; the magnitude of the 4-velocity has to be invariant.
22cuuu
4. 4-momentum, rest mass and conservation laws
In spacetime 4-momentum is mass m times 4-velocityp
u
--Under this definition, the mass must be a scalar if the 4-momentum is going to be a 4-vector.
--The mass m of an object as far as we are concerned is its rest mass, or the mass we would measure if we were at rest with respect to the object.
)ˆ,(),,,( vmmcmvmvmvmcump zyx
--Again, by switching into the rest frame of the particle, or by calculationg the magnitude we find that 4-momentum, we can show:
mcp
As with 4-velocity, it is strange but true that the magnitude of the 4-momentum does not depend on speed.
Why introduce all these 4-vectors, and in particular the 4-momentum?--all the laws of physics must be same in all uniformly moving reference frames
--only scalars and 4-vectors are truly frame-independent, relativistically invariant conservation of momentum must take a slightly different form.
--In all interactions, collisions and decays of objects, the total 4-momentum is conserved (of course we don’t consider any external force here).
--Furthermore, )ˆ,( vmmcump
c
E pWe are actually re-defining E and p to be: vmpandmcE ˆˆ2 You better forget any other expressions you learned for E or p in non-relativistic mechanics.
A very useful equation suggested by the new, correct expressions for E and p
E
cpv
2ˆˆ
Taking the magnitude-squared of p
We get a relation between m, E and pp ˆ
22
222p
c
Eppcmp
which, after multiplication by c2 and rearrangement becomes 22422 cpcmE
This is the famous equation of Einstein's, which becomes
422 cmE when the particle is at rest 0ˆ p
In the low-speed limit 1c
v
222
122
2
22
12
2
1)1(
ˆˆ2
1ˆ)1(ˆˆ
mvmcmcE
vmvc
vmvmvmp
i.e., the momentum has the classical form, and the energy is just Einstein's famous mc2 plus the classical kinetic energy mv2/2. But remember, these formulae only apply when v << c.
5. Conservation laws
qp
Summed over All the 4-momenta of all the components of the whole system before interaction
Summed over all the 4-momenta of all the components of the whole system after interaction
i
ii
i qp
For a single particle: 4-momenum before an action = that after
For a multi-particle system:
5. CollisionsIn non-relativistic mechanics collisions divide into two classes:
elastic inelastic
energy and 3-momentum are conserved.
only 3-momentum is conserved
In relativistic mechanics 4-momentum, and in particular the time component or energy, is conserved in all collisions;
No distinction is made between elastic and inelastic collisions.
m m
Before the collision After the collision
)0,0,0,();0,0,,( mcpmvmcp sm
v M’ 'v
)]0,0,,)1[(ˆ mvmcppp sm )0,0,''',''( vMcMq
By conservation of 4-momentum before and after collision, which means that the two 4-vectors are equal, component by component,
Non-relativistic theory gives: M’=2m, v’ =v/2
The ratio of these two components should provide v’/c;
21'
vvv
The magnitude of q
should be M’ c; we use
222
22
2
222222 )1(2)]1(21[]1[' mm
c
v
c
vmmM
mmM 2)1(2'
--So the non-relativistic answers are incorrect,
--the mass M’ of the final product is greater than the sum of the masses of its progenitors, 2m.
Q: Where does the extra rest mass come from? A: The answer is energy.
In this classically inelastic collision, some of the kinetic energy is lost.
But total energy is conserved. Even in classical mechanics the energy is not actually lost, it is just converted into other forms, like heat in the ball, or rotational energy of the final product, or in vibrational waves or sound travelling through the material of the ball.
22;' pqcMq
Strange as it may sound, this internal energy actually increases the mass of the product of the collision in relativistic mechanics.
The consequences of this are strange. For example, a brick becomes more massive when one heats it up. Or, a tourist becomes less massive as he or she burns calories climbing the steps of the Effiel Tower.
All these statements are true, but it is important to remember that the effect is very very small unless the internal energy of the object in questionis on the same order as mc2.
For a brick of 1 kg, mc2 is 1020 Joules, or 3 *1013 kWh, a household energy consumption over about ten billion years (roughly the age of the Universe!)
For this reason, macroscopic objects (like bricks or balls of putty) cannot possibly be put into states of relativistic motion in Earth-bound experiments.Only subatomic and atomic particles can be accelerated to relativistic speeds, and even these require huge machines (accelerators) with huge power supplies.
6. Photons and Compton scattering
i) Can something have zero rest mass?
Substitute E=pc into v = p c2/E = c
6.1 properties of photon
22422 cpcmE From E = p c (p is the magnitude of the 3-momentum)
So massless particles would always have to travel at v = c, the speed of light. Strange??
Photons, or particles of light, have zero rest mass, and this is why they always travel at the speed of light.
ii) The magnitude of a photon's 4-momentum p
pcE;00 222
222
cpc
Ecmp
but this does not mean that the components are all zero.
--The time component squared, E2/ c2, is exactly cancelled out by the sum of the space components squared,
2222 pppp zyx --Thus the photon may be massless, but it carries momentum and energy, and it should obey the law of conservation of 4-momentum.
6.2 Compton scattering.
The idea of the experiment is to beam photons of known momentum Q at a target of stationary electrons,and measure the momenta Q’of the scattered photons as a function of scattering angle.We therefore want to derive an expression for Q’ as a function of .Before the collision the 4-momenta of the photon and electron are:
);0,0,,( QQp
)0,0,0,(mcpe
after they are: )0,sin,cos,();0,sin',cos','( mvmcqQQQq e
The conservation law is ee qqpp
22 )()( ee pqqp
)(22 aqpqqppqpqqpp eeeeee
For all photons ;0pp and for all electrons2)(mcpp
Also, in this case cos'' QQQQqp
;22cmqp ee And: 22)1(2)cos1('2 cmQQ Equation (a) becomes:
But by conservation of energy, ( −1)mc is just Q−Q’, and (a − b)/ab is just 1/b−1/a, so we have what we are looking for:
)cos1(11
'
1 mcQQ
hQhhv
c
E
c
Q
'
'
1;
1
This prediction of special relativity was confirmed in a beautiful experiment by Compton (1923) and has been reconfirmed many times since by undergraduates in physics lab courses.
In addition to providing quantitative confirmation of relativistic mechanics, this experimental result is a demonstration of the fact that photons, though massless, carry momentum and energy.
Quantum mechanics tells
The energy E of a photon is related to its wave frequency by E = h
so we can rewrite the Compton scattering equation in its traditional form:
Then
)cos1(' mc
h
An elementary particle of rest mass M decays from rest into a photon and a new particle of rest mass M/2. Find its velocity.
M
M/2 u
h
For 3-momentum conservation, the particle moves in x direction, and the photon moves in –x direction.
)0,0,0,(McpM )0,0,2
,2
();0,0,,(2
MuMcp
c
hv
c
hvp Mph
;
0
02
2
0
00
0
0
2
Mu
Mc
c
hvc
hvMc
ppp phMM
)2(2
Mu
c
hv
)1(2
Mc
c
hvMc
(2) Into (1): cu
cuMc
cu
cuMcuMMcMuMc
/1
/1
212
)(
2
)(
222
2
7. Particle decay and pair production
cu 6.0Solve for u:
7.1 Particle decay:
By momentum conservation:
7.2 Pair production - gamma photon can not be converted to e- and e+
Show that the following pair production cannot occur without involvement of other particles. e-
e+
Let m be the rest mass of electrons and u, v the 3-velocities of electron and positron.
000
02
2
2
1
1
1
y
x
y
xph vm
mv
cm
mu
mu
cm
c
hvc
hv
p
)3(
)2()(
)1()(
21
21
212
yy
xx
vu
vumc
hv
mchv
Sub. (3) into (2):
)4()()(1
)(
2
1y
yxx
y
yxx v
uvu
mccu
hv
v
uvum
c
hv
Sub. (3) into (1):
)5()1()(1
)1(
2
12
y
y
y
y
v
uc
mccu
hv
v
umc
c
hv
Compare (4) and (5):)6()1()(
y
y
y
yxx v
uc
v
uvu For ux and vx < c
(6) can not be satisfied
Pair production needs an additional particle to carry off some momentum.
8 Velocity addition (revisited) and the Doppler shift
In S, a particle of mass m moves in the x-direction at speed vx, so its 4-momentum is
2
2111
1
1where)0,0,,(
cv
mvmcpx
x
In S’ moving at speed v, the 4-momentum of the particle:
2
22
1
1
cv
andc
vwith
2
'
2121
2121
2121
2121
1/
'
/
/
'
''
c
vv
vvv
cvvc
vv
c
v
cvmvmc
cmcvmv
mc
mv
x
xx
x
xx
x
xx
8 .1 Velocity addition revisited
This is a much simpler derivation than that found before.
8.2 Photon makes an angle from x axis
Q(fobs )S y
x
z
v Q’(frest )S’y’
x’
z’ );0,sin,cos,( QQQq )0,'sin','cos','(' QQQq
;
0
'sin'
'cos'
'
1000
0100
00
00
0
sin
cos
Q
Q
Q
Q
Q
Q
q
’
Equate each component on both side:
'sin'sin
'cos''cos
'cos''
QQQ
QQQ
21
cos1)'cos1('
em
emobs ffQQ
em
emobs
Q
Q
cos1
coscos
'cos1
'cos
)'cos1(
)'cos()'cos(
'cos
c
v
21
cos1
em
emobs ffDoppler effect from:
you.towardmovingissourcelightthewhen,shitedblueislightthe
1
1
1
1;0If i)
2
ememobsem fff
you.fromaway movingissourcelightthewhen,shitedredislightthe
1
1
1
1;If ii)
2
ememobsem fff
effect.Doppler e transverssmall very a predicts This
1
1;2/If ii)
2
emobsem ff
EDclassicc
vff emob
)1/(;
cv
1
1
c
v1 :cvWhen
EDclassicc
vff emob
)1/(;
cv
1
1
c
v1 :cvWhen
em
emobs
cos1
coscos
Aberration of light from:
0;1cos;0if obsobsem
obsobsem ;1cos;if
2/0;cos;2/if obsobsem
1
2
1
2
3
3
Light rays emitted by source in S’ Light rays observed in S
When v is very large so that =0.9, and cosobs =0.9, obs =26
http://www.anu.edu.au/Physics/Savage/TEE/site/tee/learning/aberration/aberration.html
a
if we want to define a 4-vector form of acceleration , or a 4-vector force , we will need to use
K
d
pdKand
d
uda
9. 4-force
umpandd
xdu
We recall the 4-velocity and 4-momentum are defined in terms of derivativeswith respect to proper time rather than coordinate time t . The definitions are
Where is spacetime position and m is rest massx
)ˆ,( Because pc
Ep
)ˆ
,( d
pd
cd
dEK
Fd
pd ˆˆ
Also, if the rest mass m of the object in question is a constant (not true if the object in question is doing work, because then it must be using up some of its rest energy!),
0
0
0)(
22
Kp
d
pdpp
d
pdd
ppd
cmpp
i.e., if the rest mass is not changing then and are orthogonal. In 3+1-dimensional spacetime, orthogonalityis something quite different from orthogonality in 3-space: it has nothing to do with 90 angles.
p
K